# Thread: Proof of span(S) is a subspace:a question

1. ## Proof of span(S) is a subspace:a question

We know the definition of subspace which is:

If $W$ is a set of one or more vectors from a vector space $V$, then $W$ is a subsace of $V$ if and only if the following conditions hold:

(a) If $u$ and $v$ are vectors in $W$, then $u + v$ is in $W$.
(b) If $k$ is any scalar and $u$ is any vector in $W$, then $ku$ is in $W$.

Also the definition of $span(S)$ is:
If $S = \{v_1, v_2, ..., v_r\}$ is a set of vectors in a vector space $V$, then the subspace $W$ of $V$ consisting of all linear combinations of the vectors in $S$ is called the space spanned by $v_1, v_2, ...,v_r$, and we say that the vectors $v_1, v_2, ..., v_r$ span $W$. To indicate that $W$ is the space spanned by the vectors in the set $S = \{v_1, v_2,..., v_r\}$, we write:

$W = span(S)$ or $W = span\{v_1, v_2,..., v_r\}$

Now the problem I'm having is how is it possible that $span(S)$ is a subspace? I can't make a connection with the definition and the proof.

Is it possible to explain the proof of why $span(S)$ is a subspace?

2. what does an element of span(S) look like? more pointedly, how do you form a "linear combination of the vj" what does that MEAN?

when you have hit upon that, you will see at once that two such linear combinations may be combined, and that BY THEIR LINEARITY, combined into a single linear combination.

here is a two-dimensional example: suppose S = {u,v}. we may regard an element of span(S) as being of the form a1u + a2v. another such element might be: b1u + b2v.

what could be more natural than (a1u + a2v) + (b1u + b2v) = (a1+b1)u + (a2+b2)v. is this not a linear combination of u and v, as well?

extend this idea to a scalar product k(a1u + a2v).

it should be clear that if S is a non-empty set, that span(S) has "one or more vectors" in it. convince yourself that for u in S, ku in span(S) implies that the 0-vector is in span(S).

can you see that -u has to be in span(S) as well? what further convincing might you need to see that span(S) has all the properties of a vector space in and of itself, and being a subset, is thus a sub-space?

3. The span, W, of a set of vectors $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is, as you say, the set of all possible linear combinations of those vectors. Now, suppose u and v are vectors in W. Then $u= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$ for some numbers $a_1, a_2, \cdot\cdot\cdot, a_n$. Also, $v= b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n$.

Then, $u+ v= (a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)+ (b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n)= (a_1+ b_1)v_1+ (a_2+ b_2)v_2+ \cdot\cdot\cdot+ (a_n+ b_n)v_n$ which is again a linear combination of $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ and so in W. Similarly, $ku= k(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)= ka_1v_1+ ka_2v_2+ \cdot\cdot\cdot+ ka_nv_n$ a linear combination of $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ and so in W.

4. I don't how I missed such a simple thing. It makes a lot of sense now. Thanks HallsofIvy and Deveno both for your explanations and extraordinary brains.

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# Prove that span(S) is a subspace of V

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